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snow_lady [41]
4 years ago
15

What is the inverse of f(x)=(x+6)2 for x≥–6 where function g is the inverse of function f? g(x)=x√−6, x≥0 g(x)=x−6−−−−√, x≥6 g(x

)=x√+6, x≥0 g(x)=x+6−−−−√, x≥−6
Mathematics
1 answer:
rosijanka [135]4 years ago
7 0

Answer:

g(x)=\sqrt{x}-6, x>=0

Step-by-step explanation:

f(x)=(x+6)^2

To find the inverse function , replace f(x) by y

y=(x+6)^2

Replace x with y and y with x

x=(y+6)^2

Solve the equation for y

take square root on both sides

+\sqrt{x} =y+6

Now subtract 6 from both sides

+\sqrt{x}-6=y

replace y with g(x)

g(x)=\sqrt{x}-6

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Find the solution to the system of equations: x + 3y = 7
tigry1 [53]

Answer:

(-2,3)

Step-by-step explanation:

Doing it the way you did and working out the problem you substituted 3 into, x+3(3)=7, 3(3) is 9 and then you subtract 9 from both sides.

7-9=-2

x=-2

Then all you have to do is put it into ordered pairs.

(x,y)

(-2,3)

3 0
3 years ago
There are 80 girls and 70 boys in the 6th Grade at a middle school. Of these students 12 girls and 9 boys write left handed. Wha
laila [671]

Answer:

About 7% of the students

Step-by-step explanation:

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4 years ago
Read 2 more answers
Evaluate each function following the specification.
telo118 [61]

Answer:

#3 a. g(-1) = 2, g(0) = 3, and g(1) = 2

b. No restrictions for all real numbers

\#4 \  a. \ h(-1) = \dfrac{1}{3}, \  h(0) =0, \  h(2) =  \infty

b. Yes, <em>x ≠ 2</em>

Step-by-step explanation:

#3 The function is given as g(x) = -x² + 3

a. From the given function, by plugging in the value of 'x' in the bracket, we have;

g(-1) = -(-1)² + 3 = -1 + 3 = 2

g(-1) = 2

g(0) = -0² + 3 = 3

g(0) = 3

g(1) = -1² + 3 = -1 + 3 = 2

g(1) = 2

g(-1) = 2, g(0) = 3, and g(1) = 2

b. The given function g(x) = -x² + 3 for finding the value of <em>g</em> can take any value of <em>x</em> which is a real number

Therefore, therefore, there are no restrictions

#4 a. The given function is given as follows;

h(x) = \dfrac{x}{x - 2}

By substitution, we get;

h(-1) = \dfrac{-1}{(-1) - 2} = \dfrac{-1}{-3} = \dfrac{1}{3}

\therefore h(-1) = \dfrac{1}{3}

h(0) = \dfrac{0}{0 - 2} = \dfrac{0}{-2} =0

\therefore h(0) =0

h(2) = \dfrac{2}{2 - 2} = \dfrac{2}{0} = \infty

\therefore h(2) =  \infty

h(-1) = \dfrac{1}{3}, \  h(0) =0, \  h(2) =  \infty

b. From the values of the function, we have that h(x) is not defined at x = 2

Therefore, there is a restriction for <em>x</em> in the function, which is <em>x ≠ 2</em>

3 0
3 years ago
PLEASE HELP!!
Bezzdna [24]

Answer:

y=-\dfrac{9}{40}x^2+\dfrac{441}{40}\\ \\y=-\dfrac{9}{160}x^2+\dfrac{1,521}{160}

Step-by-step explanation:

If a parabola has its vertex on the y-axis, then its equation is

y=ax^2+b

This parabola passes through the point R(3,9), then

9=a\cdot 3^2+b\\ \\9=9a+b

The area of the right triangle PQR is

A_{PQR}=\dfrac{1}{2}\cdot PQ\cdot QR

Find PQ and QR, if P(x_1,0),\ Q(3,0),\ R(3,9):

PQ=\sqrt{(x_1-3)^2+(0-0)^2}=|x_1-3|\\\\QR=\sqrt{(3-3)^2+(0-9)^2}=9

Now,

40.5=\dfrac{1}{2}\cdot |x_1-3|\cdot 9\\ \\90=9|x_1-3|\\ \\|x_1-3|=10\\ \\x_1-3=10\ \text{or}\ x_1-3=-10\\ \\x_1=13\ \text{or }x_1=-7

We get two possible points P_1(-7,0) and P_2(13,0).

For point P_1:\\

0=a\cdot (-7)^2+b\\ \\49a+b=0

So,

b=-49a\\ \\9=9a-49a\\ \\-40a=9\\ \\a=-\dfrac{9}{40}\\ \\b=\dfrac{441}{40}\\ \\y=-\dfrac{9}{40}x^2+\dfrac{441}{40}

For point P_2:\\

0=a\cdot (13)^2+b\\ \\169a+b=0

So,

b=-169a\\ \\9=9a-169a\\ \\-160a=9\\ \\a=-\dfrac{9}{160}\\ \\b=\dfrac{1,521}{160}\\ \\y=-\dfrac{9}{160}x^2+\dfrac{1,521}{160}

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3 years ago
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